Department of Chemistry (Japanese)2 exp ν/k B T []()hν/k B −1 2 (4.1.12) 9CEF =8. 9CEF =8. KY,...

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Transcript of Department of Chemistry (Japanese)2 exp ν/k B T []()hν/k B −1 2 (4.1.12) 9CEF =8. 9CEF =8. KY,...

19 20

19

I.

1687 J.C.

1864

J.W. 1876 1878 H.L.F.v. 1882

1897 19

1911 L.E. 1887

20

19

A.

1905 1916

19

656.5 nm 486.3

nm 434.2 nm 410.3 nm

J.J. 1885 λ

1

λ= RH

1

22−1

n2⎛

⎝ ⎜

⎠ ⎟ (4.1.1)

RH 109678 cm-1

n = 3 4 5 6 n = 7 397.1 nm n =

8 389.0 nm (4.1.1)

20 (4.1.1)

1

λ= RH

1

l2−1

n2⎛

⎝ ⎜

⎠ ⎟ (4.1.2)

l = 1 1906 l = 3 1908

l = 4 1922 l = 2

(4.1.2)

n

l

4.1

ν

1886

1900 4-1

1900 M. 1

ρ(ν) =8πν 2

c 3hν

exp hν /kBT( ) −1 (4.1.3)2

h T

kB 1

R R = NAkB (4.1.3)

A. 1905 ν hν

ν

hν hν/c

1

W 4.2

K

2 exp(x) ex x

4.2W

K = hν −W (4.1.4)

h

X A.H.

1923

ν hν

hν/c

4.3

i o m V

4.3

θ

(mV )2 =hν ic

⎝ ⎜

⎠ ⎟ 2

+hν oc

⎝ ⎜

⎠ ⎟ 2

− 2hν ic

⎝ ⎜

⎠ ⎟ hν oc

⎝ ⎜

⎠ ⎟ cosθ

=h2

c 2ν i2 + ν o

2 − 2ν iν o cosθ( ) (4.1.5)

=h2

c 2ν i −ν o( )2 + 2ν iν o 1− cosθ( )[ ]

m2V 2 = 2h2

c 2ν iν o 1− cosθ( ) (4.1.6)

1

2mV 2 = h ν i −ν o( ) (4.1.7)

m ν i −ν o( ) = h

c 2ν iν o 1− cosθ( ) (4.1.8)

V λ

c = νλ (4.1.9)

(4.1.8)

Δλ = λo − λi =h

mc1− cosθ( ) (4.1.10)

h m c

C

N

C = 3R ×N

NA

(4.1.11)

4.4 + 1000 K

3 1819

L.E. 1871

1 mol 3R

4.4 1905 W.H.

0

A. 1907

ν hν

C =N

NA

⎝ ⎜

⎠ ⎟ kB

hνkBT

⎝ ⎜

⎠ ⎟

2exp hν /kBT( )

exp hν /kBT( ) −1[ ]2 (4.1.12)

T = 0 K 0

1 3

3R

P. 1912

20

1911 E.

10-4

1800 1

r

F = –Ar/|r|3 (4.2.1)

A

ke2 =e2

4πε0 (4.2.2)

e ε0

Ae = 2.3•10-28 N m-2 G m M

GmM Ag = 2.7•10-39 N m-2 Ag << Ae

r 4

mV 2

r= k

e2

r2 (4.2.3)

V (4.2.3)

1

2mV 2 + −k

e2

r

⎝ ⎜

⎠ ⎟ = −

1

2ke2

r (4.2.4)

r

10-11 s

A

N.H.D.

1913

= h/2π

n

mVr = n (4.2.5)

(4.2.3) V r

r = n2

2

kme2 (4.2.6)

(4.2.6) n

a0 =

2

kme2 (4.2.7)

a0

52.9 pm

(4.2.6)

r = n2a0 (4.2.8)

r (4.2.4)

En = −1

2n2ke2

a0 (4.2.9)

n

Eh =

ke2

a0=k 2me4

2 (4.2.10)

Eh

n (4.2.9)

i f i > f

ΔE = Ei − E f = −1

ni2 −

1

n f2

⎝ ⎜ ⎜

⎠ ⎟ ⎟ 1

2Eh (4.2.11)

ν λ hν

1

n f2 −

1

ni2

⎝ ⎜ ⎜

⎠ ⎟ ⎟ 1

2Eh = hν = h

c

λ (4.2.12)

1

λ=

1

n f2 −

1

ni2

⎝ ⎜ ⎜

⎠ ⎟ ⎟ 1

2hcEh (4.2.13)

m mM/(m + M)

1

λ=Eh

2hc

1

1+ (m /M)1

n f2 −

1

ni2

⎝ ⎜ ⎜

⎠ ⎟ ⎟ (4.2.14)

(4.1.2)

RH =Eh

2hc

1

1+ (m /M) (4.2.15)

1923 λ

p

λ =h

p (4.3.1)

(4.2.5) mV

2πr = nλ (4.3.2)

50 kg 6 km h-1 80 N s

h/p ≈ (6.6•10-34 J s)/(80 N s) ≈ 8•10-36 m

100 V 0.1 nm

1925 W.M.

4.5 F(x) F(x ± ct)

x t = 0 F(x)

F(x − ct) c F(x + ct)

c 4.5 ψ(x, t) = F(x ± ct)

B

∂∂xψ(x, t) =

∂X∂x

dF(X)

dX=dF(X)

dX

∂∂tψ(x, t) =

∂X∂t

dF(X)

dX= ±c

dF(X)

dX (4.3.3)

∂2

∂t 2ψ(x, t) = c 2

d2F(X)

dX 2

ψ(x, t)

∂2

∂x 2ψ(x, t) −

1

c 2∂2

∂t 2ψ(x,t) = 0 (4.3.4)

5 ψ(x, t) = F(x ± ct)

F(x − ct) F(x + ct)

a F(x − ct) + bF(x + ct) (4.3.4)

5

∂2

∂x 2+∂2

∂y 2+∂2

∂z2⎛

⎝ ⎜

⎠ ⎟ ψ(x,y,z,t) −

1

c 2∂2

∂t 2ψ(x,y,z,t) = 0 (4.3.5)

(4.3.5)

∇ =∂∂x,∂∂y,∂∂z

⎝ ⎜

⎠ ⎟ (4.3.6)

∇2 =∇ •∇ =∂2

∂x 2+∂2

∂y 2+∂2

∂z2 (4.3.7)

∇2ψ −1

c 2∂2

∂t 2ψ = 0 (4.3.8)

A ν λ x

ψ = Asin 2πx

λ−νt

⎝ ⎜

⎠ ⎟

⎣ ⎢

⎦ ⎥ (4.3.9)

(4.3.4) c λ ν

c = νλ (4.3.10)

x0 x0

∂∂tψ = −2πνAcos 2π

x0λ−νt

⎝ ⎜

⎠ ⎟

⎣ ⎢

⎦ ⎥ (4.3.11)

ν 2A2

ψ =ψ0(x,y,z)exp(i2πνt) (4.3.12)

B

exp(iθ) = eiθ = cosθ + isinθ (4.3.13)

∂2

∂t 2ψ = (i2πν)2ψ0(x,y,z)exp(i2πνt) = −(2πν )

2ψ (4.3.14)

(4.3.8)

∇2ψ +4π2ν 2

c 2ψ = 0 (4.3.15)

∇2ψ +4π2

λ2ψ = 0 (4.3.16)

E U(x, y, z)

E =1

2mv 2 +U =

1

2mp2 +U (4.3.17)

p2 = 2m[E – U(x, y, z)] p λ

(4.3.1) (4.3.16) λ

∇2ψ +8mπ2

h2[E −U(x,y,z)]ψ = 0 (4.3.18)

2

2m∇2 +U(x,y,z)

⎣ ⎢

⎦ ⎥ ψ = Eψ (4.3.19)

(4.3.18) (4.3.19) E. 1926

–ke2/r

W.

1925

(4.3.19)

H

Hψ = Eψ (4.3.20)

1969

I 1983

D.A. J.D.

1999

© K. Saito, 04/14/2016